2

Given: $ S = \sum_{i=1}^{n-1}{i! \over n!} $

How would I find the sum for an arbitrarily large $n$ ?

Example: $n=5$

$ S = \frac{1!}{5!} + \frac{2!}{5!} + \frac{3!}{5!} + \frac{4!}{5!} = 0.275 $

dariusc
  • 121
  • 2
    Closely related to this question which may help you http://math.stackexchange.com/questions/227551/sum-k-1-2-3-cdots-n-is-there-a-generic-formula-for-this – Carl Christian Jan 31 '16 at 19:15
  • I recommend calling such sequences $S_n$ or perhaps $S(n)$, since $S$ is dependent on $n$, and so $S$ is not actually a number (more like a function in $n$ or a sequence). I'm not saying it's wrong what you're doing, it's simply a little unusual. Also, it seems like for large $n$, this sequence tends to $0$, with maxima at $n=2$ and $n=3$ where your sequence equals $\frac12$ –  Jan 31 '16 at 19:20
  • I'm not a mathematician, I'm a programmer, so forgive me for any unusual presentation. What I'm trying to accomplish is how to generate a sum for the series for a large $n$, say.. $n=10110$, which would quickly overflow if not handled properly – dariusc Jan 31 '16 at 19:25
  • 1
    That you're a programmer, is in my opinion very useful context. To calculate this sum straight-forward without having to worry about an overflow, note that $\frac{i!}{n!}=\frac{1}{(i+1)(i+2)\cdots (n-1)n}$. Other than that, as you can see in the related post linked to by Carl Christian, there's probably no real smart way to compute this sum exactly (unless you can use integrals) - you could of course consider trying an approximation of the factorial, Stirling's approximation –  Jan 31 '16 at 19:29
  • That just put me on a good path to solving it, thanks! – dariusc Jan 31 '16 at 19:32

0 Answers0